A beam of light consisting of two wavelengths `650 nm and 520 nm`, is used to obtain interference fringes in a Young's double slit experiment. (a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength `650 nm`. (b) What is the least distance from the central maximum, where the bright fringes due to both the wavelength coincide ?

Explanation

wavelength of the light beam `lamda _(1) = 650 nm`
Wavelength of another beam ,`lamda _(2) =520 nm`
Distance of the slits from the scron =D
Distance between the two slits =d
(a) Distance of the `n^(th)` birght frinage on the screen from the central maximum is given by the relation ,
`x=nlamda_(1)((D)/(d))`
for third bright fringe n=3
`therefore x=3 3xx650 (D)/(d) =1950((d)/(d)) nm `
(b) Let the `n^(th)` bright due wavelength `lamda_(2)` and `(n-1 )^(th)` bright fringe due to wavelength `lamda_(1)` coincide on the screen . we can equate the consitions for bright fringes as :
`nlamda_(2)=(n-1)lamda_(1)`
520 n=650 n-650
650 =130 n
`therefore `n=5
Hence , the distance from central maximum can be obtained by the realation:
`x=nlamda_(2) (d)/(d)`
` =5 xx520 (D)/(d)nm`
Note : the value of d and D are not given in the question .